HDU 4722 Good Numbers(找规律)
2015-05-07 15:27
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Description
If we sum up every digit of a number and the result can be exactly divided by 10, we say this number is a good number.
You are required to count the number of good numbers in the range from A to B, inclusive.
Input
The first line has a number T (T <= 10000) , indicating the number of test cases.
Each test case comes with a single line with two numbers A and B (0 <= A <= B <= 10 18).
Output
For test case X, output "Case #X: " first, then output the number of good numbers in a single line.
Sample Input
Sample Output
Hint
If we sum up every digit of a number and the result can be exactly divided by 10, we say this number is a good number.
You are required to count the number of good numbers in the range from A to B, inclusive.
Input
The first line has a number T (T <= 10000) , indicating the number of test cases.
Each test case comes with a single line with two numbers A and B (0 <= A <= B <= 10 18).
Output
For test case X, output "Case #X: " first, then output the number of good numbers in a single line.
Sample Input
2 1 10 1 20
Sample Output
Case #1: 0 Case #2: 1
Hint
The answer maybe very large, we recommend you to use long long instead of int.
#include <iostream> using namespace std; typedef long long ll; int is(ll n) { ll m=n,i; for(i=n/10*10;i<=m;i++) { ll sum=0,t=i; while(t!=0) { sum+=t%10; t/=10; } if(sum%10==0) return 1; } return 0; } ll get(ll n) { if(is(n)==1) return n/10+1; return n/10; } int main() { int t,cnt=1; cin>>t; while(t--) { ll a,b; cin>>a>>b; cout<<"Case #"<<cnt++<<": "<<get(b)-get(a-1)<<endl; } return 0; }
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